a: \(AB=\sqrt{\left(3-0\right)^2+\left(-2-2\right)^2+\left(1-1\right)^2}=5\)
\(AC=\sqrt{\left(-2-0\right)^2+\left(5-2\right)^2+\left(7-1\right)^2}=7\)
\(BC=\sqrt{\left(-2-3\right)^2+\left(5+2\right)^2+\left(7-1\right)^2}=\sqrt{110}\)
Chu vi tam giác ABC là:
\(C_{ABC}=AB+AC+BC=12+\sqrt{110}\)
b: Xét ΔABC có \(cosBAC=\dfrac{AB^2+AC^2-BC^2}{2\cdot AB\cdot AC}=\dfrac{5^2+7^2-110}{2\cdot5\cdot7}=\dfrac{-36}{70}=-\dfrac{18}{35}\)
=>\(\widehat{BAC}\simeq121^0\)
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